Problem: $n$ coins are simultaneously flipped.  The probability that at most one of them shows tails is $\frac{3}{16}$.  Find $n$.
Answer: Since each coin has 2 possible outcomes, there are $2^n$ possible outcomes for the $n$ coins.  The number of outcomes in which the number of tails is 0 or 1 is $\binom{n}{0}+\binom{n}{1}=1+n$.  So the probability of having at most one tail is $\dfrac{1+n}{2^n}$.  Therefore, we must solve the equation $$ \frac{1+n}{2^n} =\frac{3}{16}. $$ We can check (simply by plugging in values of $n$) that if $1 \leq n \leq 5$, then $n=5$ is the only solution.  Now we show that $n\geq 6$ cannot be a solution to the equation.  Observe that $n\geq 6$ implies $n<2^{n-3}$, thus \[\frac{1+n}{2^n}<\frac{1+2^{n-3}}{2^n}=\frac{1}{2^n}+\frac{1}{8}<\frac{1}{16}+\frac{1}{8}=\frac{3}{16}.\] So there are $\boxed{5}$ coins.